IGCSE Mathematics - Additional (0606) - 2021 Edition
3.06 Absolute value equations and inequalities
Lesson

Absolute value equations

Like other functions, a function that has the absolute value symbols in its definition becomes an equation when its function value is specified. The equation may then be solved. We may get $2$2, $1$1 or $0$0 solutions. Consider under when we will have the different number of solutions using the graph below.

Number of solutions for absolute value equations

Definition of absolute value

For any real numbers $a$a and $b$b, where $b\ge0$b0, if $\left|a\right|=b$|a|=b, then $a=b$a=b or $a=-b$a=b.

Worked examples

EXAMPLE 1

Solve the equation $2|x-1|+3=21$2|x1|+3=21.

Think: We can start solving this like a regular linear equation until we get to the absolute value sign. To be sure of finding all the solutions to such an equation, we must look at both sides of the absolute value function.

Do:

 $2\left|x-1\right|+3$2|x−1|+3 $=$= $21$21 (State the original equation) $2\left|x-1\right|$2|x−1| $=$= $18$18 (Subtract $3$3 from both sides) $\left|x-1\right|$|x−1| $=$= $9$9 (Divide both sides by$2$2) $x-1$x−1 $=$= $9$9 Split into the two cases $x-1$x−1 $=$= $-9$−9 $x$x $=$= $10$10 (Add $1$1to both sides) $x$x $=$= $-8$−8

The existence of these solutions can be seen in the following graphical representation.

EXAMPLE 2

Solve the equation $|2x+7|=|x-5|$|2x+7|=|x5|.

Think: Since everything is inside absolute value signs, we need to split the equation into two cases right away. The first case is when both expressions have the same sign, and the other is when they have the opposite sign.

Do:

 $2x+7$2x+7 $=$= $x-5$x−5 Split into the two cases $2x+7$2x+7 $=$= $-(x-5)$−(x−5) $x$x $=$= $-12$−12 Solve both equations $2x+7$2x+7 $=$= $-x+5$−x+5 $3x+7$3x+7 $=$= $5$5 $3x$3x $=$= $-2$−2 $x$x $=$= $-\frac{2}{3}$−23​

Note: The reason that there are only two equations is because if we take the positive and negative versions of the expressions inside the absolute value signs we get two sets of equivalent equations.

•  If we make both expressions positive: $2x+7=x-5$2x+7=x5, this equation will have the same solution as if we make both expressions negative: $-(2x+7)=-(x-5)$(2x+7)=(x5). So rather than solve both equations and get the same answer, we only need to solve one.
• If we make the left expression positive and the right expression negative: $2x+7=-(x-5)$2x+7=(x5), this equation will have the same solution as if we make the left expression negative and the right expression positive: $-(2x+7)=x-5$(2x+7)=x5. So rather than solve both equations and get the same answer, we only need to solve one.
• This is because when you multiply an equation with a negative you get an equivalent equation.

Practice questions

Question 2

Solve the equation $9\left|x\right|=4$9|x|=4.

Write both solutions as equations on the same line separated by a comma.

Question 3

Consider the equation $\left|5x\right|=15$|5x|=15.

1. On the same set of axes, graph the functions $y=\left|5x\right|$y=|5x| and $y=15$y=15.

2. Hence determine the solutions to the equation $\left|5x\right|=15$|5x|=15. Write the solutions on the same line, separated by a comma.

Question 4

Solve $\left|4x-8\right|+1=13$|4x8|+1=13.

Write both solutions as equations on the same line separated by a comma.

Absolute value inequalities

Inequalities of the form $\left|x\right||x|<b or$\left|x\right|\le b$|x|≤b To say$x\le4$x4 is to say that$x$x can be any number positive or negative that is not greater than four. On a number line we draw$x\le4$x4 as such: If the notation is changed slightly to$|x|\le4$|x|4, then we are restricting$x$x to be any number between$-4$4 and$4$4, inclusive. On a number line we draw$|x|\le4$|x|4 as such: You could check that if$x$x is a number less than$-4$4, then the absolute value of the number is greater than$4$4. It follows that a statement like$|x|\le4$|x|4 is equivalent to a pair of inequalities that can be written$-4\le x\le4$4x4. See the different representations below: Absolute Value Inequality Interval Notation Number Line$\left|x\right|\le4$|x|4$-4\le x\le4$4x4$\left[-4,4\right]$[4,4]$\left|x\right|<2$|x|<2$-22<x<2 $\left(-2,2\right)$(2,2)

To solve an inequality (sometimes called an inequation) means to find out the range or ranges of values of the variable for which the statement is true.

Above we looked at $|x|\le4$|x|4, and said this was the same as restricting $x$x to be any number between $-4$4 and $4$4, inclusive.

What about something like $\left|x\right|\ge2$|x|2? Well, any number greater than or equal to $2$2 will satisfy this equation, but also any values less than or equal to $-2$2. This means that this single expression is represented by two disjoint intervals as show below.

See the different representations below:

Absolute Value Inequality Interval Notation Number Line
$\left|x\right|\ge2$|x|2 $x\le-2$x2 or $x\ge2$x2 $\left(-\infty,-2\right]\cup\left[2,\infty\right)$(,2][2,)
$\left|x\right|>2$|x|>2 $x<-2$x<2 or $x>2$x>2 $\left(-\infty,-2\right)\cup\left(2,\infty\right)$(,2)(2,)

Worked examples

EXAMPLE 3

Solve $\left|x+2\right|>5$|x+2|>5.

Think: It is common practice to consider what happens to the statement in $2$2 parts.  The positive side and the negative side.

Do:

Positive side means to consider and solve $+(x+2)>5$+(x+2)>5

 $+(x+2)$+(x+2) $>$> $5$5 (original positive inequality) $x+2$x+2 $>$> $5$5 (remove unnecessary brackets) $x$x $>$> $3$3 (subtract $2$2 from both sides)

Negative side means to consider and solve $-(x+2)>5$(x+2)>5

 $-(x+2)$−(x+2) $>$> $5$5 (original positive inequality) $x+2$x+2 $<$< $-5$−5 (divide both sides by $-1$−1, remember this flips the sign) $x$x $<$< $-7$−7 (subtract $2$2 from both sides)

This means that for values of $x$x, that are either greater then $3$3 or less than $-7$7, then the inequality $|x+2|>5$|x+2|>5 will hold true.

This graph shows the solution set.  Spot check values inside or outside the range to check.

Alternative approach for Question 3

Another way to think about the previous example is via translations.

Consider first the solution on a number line for $|x|>5$|x|>5.  We would know that this means all values of $x$x on the line that are a distance of greater than $5$5 from zero.  It would look like this.

The effect of the $+2$+2, is a horizontal translation of $2$2 units to the left.  Resulting in our final solution:

EXAMPLE 4

Solve the inequality $|3x+7|\le|x+9|$|3x+7||x+9|.

Think: For absolute value inequalities, with absolute values on both sides, do the following process:

1. Replace the inequality sign with an equals sign and solve the absolute value equation using two cases as in the above examples.
2. Substitute values of $x$x on either side of the $x$x-values from step $1$1, to determine which values satisfy the original inequality.
3. Write the final solution as an inequality.

Do: Replace the inequality sign with an equals sign and solve the two cases:

 Case 1: $3x+7$3x+7 $=$= $x+9$x+9 Case 2: $3x+7$3x+7 $=$= $-(x+9)$−(x+9) $2x$2x $=$= $2$2 $3x+7$3x+7 $=$= $-x-9$−x−9 $x$x $=$= $1$1 $4x$4x $=$= $-16$−16 $x$x $=$= $-4$−4

Now we must check values on each side of these two values to determine our final solution:

• $x\le-4:$x4:

If  $x=-5:\quad|3x+7|=8,|x+9|=4.$x=5: |3x+7|=8,|x+9|=4. So the inequality is false.

• $-4\le x\le1:$4x1:

If  $x=0:\quad|3x+7|=7,|x+9|=9.$x=0: |3x+7|=7,|x+9|=9. So the inequality is true: $|3x+7|\le|x+9|$|3x+7||x+9|

• $x\ge1:$x1:

If  $x=2:\quad|3x+7|=13,|x+9|=1.$x=2: |3x+7|=13,|x+9|=1. So the inequality is false.

Therefore our final solution is: $-4\le x\le1$4x1.

Practice questions

Question 5

Rewrite the inequality $\left|x\right|<2$|x|<2 without using absolute values.

Question 6

Rewrite the inequality $\left|x\right|\ge5$|x|5 without using absolute values.

Question 7

Solve $\left|4x-8\right|+1=13$|4x8|+1=13.

Write both solutions as equations on the same line separated by a comma.

Outcomes

0606C3.1

Solve graphically or algebraically equations of the type |ax + b| = c (c ⩾ 0) and |ax + b| = |cx + d|.

0606C3.2

Solve graphically or algebraically inequalities of the type |ax + b| > c (c ⩾ 0), |ax + b| ⩽ c (c > 0) and |ax + b| ⩽ |cx + d|.